If $p^k m^2$ is an odd perfect number with special prime $p$, then under what other conditions on $\sigma(p^k)/2$ does $k=1$ follow?

Let $$N = p^k m^2$$ be an odd perfect number with special prime $$p$$ satisfying $$p \equiv k \equiv 1 \pmod 4$$ and $$\gcd(p,m)=1$$.

Descartes (1638), Frenicle (1657), and subsequently [Sorli (2003) - Conjecture 3, Chapter 5 on page 89] conjectured that $$k=1$$ always holds for an odd perfect number $$N = p^k m^2$$.

It is fairly easy to show that if $$\sigma(p^k)/2$$ is prime, then $$k = 1$$.

Moreover, Broughan, Delbourgo, and Zhou (2013) show that if $$\sigma(p^k)/2$$ is a square, then $$k = 1$$.

An interesting scenario holds when $$\sigma(p^k)/2$$ is squarefree. Indeed, this assumption implies that $$H = \gcd(m^2,\sigma(m^2)) = \frac{m^2}{\sigma(p^k)/2} = G \times J^2$$ is not squarefree, where $$G$$ and $$J$$ are defined as $$G = \gcd(\sigma(p^k),\sigma(m^2)) = \dfrac{\bigg(\gcd(\sigma(p^k)/2,m)\bigg)^2}{\sigma(p^k)/2}$$ and $$J = \dfrac{m}{\gcd(\sigma(p^k)/2,m)}.$$ (For the case under consideration, that $$\sigma(p^k)/2$$ is squarefree, we do in fact have $$G = \sigma(p^k)/2$$ and $$J = \dfrac{m}{\sigma(p^k)/2}$$ since $$\sigma(p^k)/2 \mid m^2$$, and therefore $$\sigma(p^k)/2 \mid m$$, holds in general.)

Here is my initial question:

FIRST INQUIRY

Can you show that $$\sigma(p^k)/2$$ is squarefree likewise implies that $$k=1$$?

The reason for this inquiry is because I currently know that $$k=1$$ likewise implies that $$H$$ is not squarefree.

LAST INQUIRY

If $$p^k m^2$$ is an odd perfect number with special prime $$p$$, then under what other conditions on $$\sigma(p^k)/2$$ does $$k=1$$ follow?

This is a partial response, as it does not directly answer the original question that was asked. Additionally, what follows are actually some remarks that would be too long to fit in the Comments section.

This answer builds on the results in this MSE question from January 03, 2023, and shows that $$H = \frac{\sigma(m^2)}{p^k}$$ is not a square, if $$\sigma(p^k)/2$$ is squarefree.

Throughout this paper, we implicitly rely on the simple equality $$\sigma(m^2) = \frac{2p^k m^2}{\sigma(p^k)}. \tag{1}$$ Unfortunately, this seems to introduce fractions. To avoid that, we can use prime factorizations, as follows. Write the prime factorization of $$m$$ as $$m = {q_1}^{a_1} \cdots {q_n}^{a_n},$$ for some unique odd primes $$3 \leq q_1 < \ldots < q_n$$, and for some positive integer exponents $$a_1, \ldots, a_n$$. Since both sides of $$(1)$$ are integers, and since $$p \equiv k \equiv 1 \pmod 4$$ with $$p$$ prime, we know that $$\sigma(p^k) = 2 {q_1}^{b_1} \cdots {q_n}^{b_n}$$ for some nonnegative integers $$0 \leq b_i \leq 2a_i$$. Thus, we have $$\sigma(m^2) = p^k {q_1}^{2a_1 - b_1} \cdots {q_n}^{2a_n - b_n}.$$

Suppose that $$\sigma(p^k)/2$$ is squarefree. This assumption implies that $$\sigma(p^k) = 2 {q_1}^{b_1} \cdots {q_n}^{b_n}$$ for $$0 \leq b_i \leq 1$$.

Since $$\frac{\sigma(p^k)}{2} \geq \frac{p^k + 1}{2} \geq 3,$$ then $$b_i = 1$$ for at least one $$i$$.

This means that $$H = \frac{\sigma(m^2)}{p^k} = {q_1}^{2a_1 - b_1} \cdots {q_n}^{2a_n - b_n}$$ where $$2a_i - b_i = 2a_i - 1 \equiv 1 \pmod 2$$ for at least one $$i$$.

Consequently, $$H$$ is not a square, if $$\sigma(p^k)/2$$ is squarefree.

• In fact, by the contrapositive, if $$H = \frac{\sigma(m^2)}{p^k} = \frac{m^2}{\sigma(p^k)/2}$$ is a square, then $\sigma(p^k)/2$ is also a square (and therefore, is not squarefree). Commented Jan 5, 2023 at 10:05